An open rectangular tank is made of steel plate of volume 1440cm cube it height is twice its width if the depth of the tank is 4 m less than width. Find its length,width and height.

Answer 1

Let's use variables to represent the dimensions of the tank. Let the width be x cm, the height be 2x cm, and the depth be (x-4) cm.

To find the volume of the tank, which is given as 1440 cm^3, we use the formula: volume = length × width × height.

1440 = length × x × 2x × (x-4)

Simplifying the equation:

1440 = 2x^3 (x-4)

Dividing both sides by 2:

720 = x^3 (x-4)

Expanding the expression:

720 = x^4 - 4x^3

Rearranging the equation:

x^4 - 4x^3 - 720 = 0

This equation is a quartic equation, but it can be factored:

(x - 12)(x + 12)(x - 6)(x + 6) = 0

We can solve for the possible values of x:

x - 12 = 0 --> x = 12
x + 12 = 0 --> x = -12 (rejected since dimensions cannot be negative)
x - 6 = 0 --> x = 6
x + 6 = 0 --> x = -6 (rejected since dimensions cannot be negative)

Therefore, the width of the tank can be 12 cm or 6 cm. However, since the height is stated to be twice the width, we choose the width to be 6 cm.

Width = 6 cm
Height = 2 * Width = 2 * 6 cm = 12 cm
Depth = Width - 4 = 6 cm - 4 cm = 2 cm

Therefore, the dimensions of the tank are Length = arbitrary, Width = 6 cm, Height = 12 cm, and Depth = 2 cm.